Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const char letters[] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i',
                        'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r',
                        's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
const int dx[] = {0, -1, 0, 1, -1, -1, 1, 1};
const int dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
int inline max3(int a, int b, int c) { return max(a, max(b, c)); }
const int N = 105;
int n, m;
int grid[N][N];
int r[N], c[N], ansr[N], ansc[N];
int main() {
  int st = 1000000100, sum = 1000000100, cnt;
  bool f;
  scanf("%d", &n);
  scanf("%d", &m);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) scanf("%d", &grid[i][j]);
  for (int i = 0; i < m; ++i) st = min(st, grid[0][i]);
  for (int i = 0; i <= st; ++i) {
    cnt = r[0] = i, f = 0;
    for (int j = 0; j < m; ++j) c[j] = grid[0][j] - i, cnt += c[j];
    for (int k = 1; k < n; ++k) {
      int tst = grid[k][0] - c[0];
      if (tst < 0) {
        f = 1;
        break;
      }
      for (int z = 1; z < m; ++z)
        if (grid[k][z] - c[z] != tst) {
          f = 1;
          break;
        }
      if (f) break;
      r[k] = tst, cnt += tst;
    }
    if (f) continue;
    if (cnt < sum) {
      sum = cnt;
      for (int i = 0; i < n; ++i) ansr[i] = r[i];
      for (int i = 0; i < m; ++i) ansc[i] = c[i];
    }
  }
  if (sum == 1000000100) sum = -1;
  printf("%d\n", sum);
  if (sum != -1) {
    for (int i = 0; i < n; ++i)
      if (ansr[i])
        for (int j = 0; j < ansr[i]; ++j) printf("row %d\n", i + 1);
    for (int i = 0; i < m; ++i)
      if (ansc[i])
        for (int j = 0; j < ansc[i]; ++j) printf("col %d\n", i + 1);
  }
  return 0;
}